← Unit 3: Gravity and electromagnetism
Topic 1: Gravity and motion
Apply the relationships for uniform circular motion, including centripetal acceleration a = v^2/r, centripetal force F = m v^2 / r, period T = 2 pi r / v, and the geometry of banked curves and conical pendulums
A focused answer to the QCE Physics Unit 3 dot point on uniform circular motion. Defines centripetal acceleration, identifies the real forces that supply centripetal force in common contexts (string tension, friction, normal-force component, gravity), and works the banked curve and conical pendulum geometries that QCAA expects in IA1 and IA2.
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What this dot point is asking
QCAA wants you to apply the kinematic and dynamic relationships for an object moving in a circle at constant speed, identify the real force (or component of a real force) supplying the centripetal force in a given context, and solve geometry problems for banked curves and conical pendulums. The dot point appears in IA1 short response, often as a banked-curve stimulus, and is a popular IA2 design (conical pendulum or rotating-mass setup).
The answer
Uniform circular motion in one paragraph
An object moving at constant speed around a circle of radius has changing velocity (because direction is changing). The acceleration is directed toward the centre of the circle (centripetal):
By Newton's second law, this requires a net inward force:
Centripetal force is not a separate kind of force. It is the role played by whichever real force (or component of one) points toward the centre. In different contexts:
| Context | Source of IMATH_7 |
|---|---|
| Mass on a string in a horizontal circle | Tension (horizontal component for a conical pendulum) |
| Car on a flat bend | Static friction between tyres and road |
| Car on a banked curve (frictionless) | Horizontal component of the normal force |
| Satellite in orbit | Gravitational attraction toward the central body |
| Charged particle in a magnetic field | Magnetic Lorentz force IMATH_8 |
Period and frequency
For one complete revolution the object travels a distance at speed , so the period is:
and the frequency is . Angular speed is , so is an equivalent form.
Banked curves
On a curve banked at angle to the horizontal, the normal force is perpendicular to the road surface. Resolving:
- Vertical: (no vertical acceleration).
- Horizontal: (centripetal).
Dividing eliminates and :
This is the design speed of the banked curve. At that speed, no friction is required to keep the vehicle on the curve. Above the design speed, friction must act down the slope (toward the inside of the curve); below it, friction must act up the slope.
Try it: Banking angle calculator. Enter speed and radius to get the design angle and the friction required at any other speed.
Conical pendulum
A mass swung in a horizontal circle on a string at angle from the vertical has tension components:
- Vertical: .
- Horizontal: .
The relationship between angle and speed is identical to the banked curve: . The radius of the horizontal circle is where is the string length.
The period of a conical pendulum:
A classic IA2 measures period as a function of cone angle (or string length) and linearises against (or ) to extract .
Worked example: car on a flat bend
A 900 kg car rounds a flat bend of radius 30 m at 12 m/s. The required centripetal force is N, supplied entirely by static friction.
Try it: Centripetal force calculator. Enter mass, speed and radius for and .
How this appears in IA1 and IA2
IA1 data test. Expect a banked-curve diagram with the bank angle and a speed, asked to determine whether the vehicle stays on the curve, to find the design speed, or to extract the friction coefficient required at a non-design speed. Alternatively, a conical pendulum stimulus with multiple period measurements at different angles.
IA2 student experiment. Common designs: a conical pendulum with mass, period and angle measured to test ; a rotating-mass setup on a horizontal turntable with friction varied; a ball-in-a-cone or marble-in-a-bowl rolling at different heights. Strong reports linearise the relationship before fitting (e.g. vs slopes give ) and propagate uncertainty cleanly.
Common traps
Treating centripetal force as a new force. It is the inward-pointing role played by an existing real force. Always state which real force (tension, friction, gravity, normal-component) supplies it.
Pointing tangentially or outward. Centripetal force is always directed toward the centre of the circle. The "centrifugal" force students sometimes invoke is a pseudo-force that appears only in the rotating frame; not used in QCAA solutions.
Mixing up , and . and . Substitute consistently.
Banking with mass dependence. The design angle does not depend on . If appears in your final answer for the bank angle, you have made an error.
Treating uniform circular motion as zero acceleration. Constant speed does not mean constant velocity. The acceleration is non-zero because the direction is changing.
In one sentence
In uniform circular motion an object travels at constant speed around radius with centripetal acceleration directed toward the centre, supplied by whichever real force points inward (tension, friction, normal-component, gravity, Lorentz), and banked-curve and conical-pendulum geometries both reduce to .
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2023 QCAA-style6 marksA 1200 kg car drives around a flat (unbanked) circular bend of radius 45 m at 14 m/s. (a) Calculate the centripetal acceleration and the centripetal force required. (b) State which real force supplies this. (c) Calculate the minimum coefficient of static friction required between the tyres and the road. (d) If the road is now banked at an angle so that no friction is required at 14 m/s, calculate this angle. Use g = 9.8 m/s^2.Show worked answer →
A 6-mark answer needs the centripetal kinematics, the force identification, the friction calculation, and the bank-angle calculation.
(a) Centripetal acceleration and force.
.
(directed horizontally toward the centre).
(b) Source of the force. On a flat road, the only horizontal force on the car is the static friction between the tyres and the road, directed toward the centre of the bend.
(c) Minimum coefficient. Static friction must supply the entire :
, so .
(d) Bank angle for no friction. Resolve the normal force into a vertical component balancing gravity () and a horizontal component supplying centripetal force (). Dividing:
, so .
Markers reward the explicit centre-pointing direction of , the identification of friction as the supplying force, the inequality on , and the derivation of from the vertical and horizontal force balances.
2022 QCAA-style4 marksA 0.50 kg mass is swung in a horizontal circle of radius 0.80 m on a string that makes an angle of 30 degrees with the vertical (a conical pendulum). Calculate the tension in the string and the speed of the mass. Use g = 9.8 m/s^2.Show worked answer →
The string tension has a vertical component balancing gravity and a horizontal component supplying centripetal force toward the vertical axis of the cone.
Tension. From the vertical balance:
.
Speed. Horizontal component supplies centripetal force:
IMATH_4
IMATH_5
IMATH_6
.
Markers reward identifying the two components of tension and their roles, the vertical balance giving , and the horizontal balance giving , with consistent units throughout.
Related dot points
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A focused answer to the QCE Physics Unit 3 dot point on projectile motion. Resolves initial velocity into components, applies the constant-acceleration equations to each axis independently, and works the level-ground range and cliff-drop standards QCAA uses in IA1 stimulus and EA Paper 2.
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A focused answer to the QCE Physics Unit 3 dot point on orbital motion. Derives orbital speed from setting gravitational force equal to centripetal force, applies Kepler's third law to satellites and planets, and works the kinetic and gravitational potential energies of a circular orbit with the standard QCAA geostationary-satellite example.